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Motional emf in Rotating a Conducting Rod in a Uniform Magnetic Field

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Estimated time: 4 minutes
CISCE: Class 12

Derivation

A conducting rod of length L rotates about one end in a uniform magnetic field B, with constant angular velocity ω, perpendicular to the field.

Step 1: Area swept in time dt: In a small time dt, the rod sweeps an area:

  • dA = \[\frac {1}{2}\]L2

Step 2: Rate of change of flux:

  • \[\frac {dΦ}{dt}\] = B\[\frac {dA}{dt}\] = \[\frac {1}{2}\]BωL2

Step 3: Applying Faraday's Law:

  • ∣V∣ = \[\frac {1}{2}\]BωL2
CISCE: Class 12

Special Case: Rotation About the Midpoint

If the rod rotates about an axis through its midpoint, perpendicular to its length, the potential difference between its two ends is zero.

  • VO − VA = \[\frac {1}{8}\]BωL2
CISCE: Class 12

Direction of Higher Potential

Rotation Direction Higher Potential Point
Anticlockwise Midpoint is at higher potential
Clockwise End of the rod is at higher potential
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